Is an orthonormalized countable complete set of linearly independent vectors in a Hilbert Space also complete?

Question

Suppose that we have a Hilbert space, $H$, in which we can obtain a countable set of linearly independent vectors, $W = \{ w_1, \dots, w_n, \dots \}$, that is complete, in the following sense: for every vector $f \in H$, there exists a countable set of scalars $\{ a_1, \dots, a_n, \dots \}$ (say that the underlying field of $H$ is $\mathbb{R}$, for simplicity) such that, as $n \to \infty$, $$\left\| f - \sum_{k=1}^{n} a_k w_k \right\| \to 0,$$ where $\left\| \cdot \right\|$ is the norm induced by the inner product in $H$.

Now, suppose that we apply the Gram-Schmidt orthonormalization process to the set $W$. This produces a countable set, $W'$, which is composed solely of orthogonal vectors.

My question is: Does the set $W'$ also constitute a complete set of vectors in $H$, as it was the case with the set $W$? If not, can someone provide a counterexample?

Thank you very much.

Answer 1

Yes. If $f$ is orthogonal to each of the vectors in the orthonormal set you have obtained then it is also orthogonal to each $w_k$. From the hypothesis it would the follow that $f$ is orthogonal to itself, so $f=0$. This implies that your orthonormal set is complete.

Answer 2

If $W=\{ w_1,w_2,\cdots\}$ is a countable set of independent vectors, and $\{ w_1',w_2,',\cdots,w_n'\}$ is an orthonormal set obtained by Gram-Schmidt from $\{ w_1,w_2,\cdots,w_n\}$ for each $n=1,2,3,\cdots$, then $$ \left\|f-\sum_{k=1}^{n}\langle f,w_k'\rangle w_k'\right\| \le \left\|f-\sum_{k=1}^{n}\alpha_k w_k \right\| $$ for all choices of scalars $\{\alpha_k\}_{k=1}^{n}$. Therefore, if $f$ is in the closure of the subspace generated by $\{w_k\}_{k=1}^{\infty}$, it follows that $\sum_{k=1}^{n}\langle f,w_k'\rangle w_k'$ converges to $f$ as $n\rightarrow\infty$.